Homework 1.
5.1. First notice that the sequence
max
n≤t≤n+1
B(t) − B(n)
n = 1, 2, · · ·
is independent. By reflection principle,
max
n≤t≤n+1
d
B(t) − B(n) = max B(n + t) − B(n) = max B(t) = |B(1)|
0≤t≤1
0≤t≤1
Consequently, for any θ > 0,
o
n
o
n
p
p
P
max
B(t) − B(n) ≥ θ 2 log n = P |B(1)| ≥ θ 2 log n
n≤t≤n+1
n
o
− θ2 +o(1)
2
(n → ∞)
= exp − θ + o(1) log n = n
Thus,
∞
n
X
P
n=1
max
n≤t≤n+1
By Borel-Cantelli lemma,
o
p
B(t) − B(n) ≥ θ 2 log n
B(t) − B(n)
√
lim sup max
2 log n
n→∞ n≤t≤n+1
(
(
=∞
θ<1
<∞
θ>1
≥θ
θ<1
≤θ
θ>1
a.s.
Letting θ → 1 leads to the conclusion.
5.2. Write B(t) = B1 (t), · · · , Bd (t) , where B1 (t), · · · , Bd (t) are 1-dimensional
independent Brownian motions. Clearly, |B(t)| ≥ B1 (t). Therefore, by the law of the
iterated logarithm for 1-dimensional Brownian motion (Theorem 5.1),
lim sup √
t→∞
|B(t)|
B1 (t)
≥ lim sup √
=1
2t log log t
2t log log t
t→∞
It remains to establish the upper bound.
(b1 , · · · , bd ) in Rd , write
a·b=
d
X
j=1
For each a = (a1 , · · · , ad ) and b =
aj bj and |b| =
d
nX
j=1
Here we use the fact that
|x| = sup a · x
|a|=1
1
a.s.
x ∈ Rd
b2j
o1/2
Given a ǫ > 0, we have
{x ∈ Rd ; |x| = 1} ⊂
[
{x ∈ Rd ; a · x > 1 − ǫ}
|a|=1
By finite cover theorem, there are integer m = m(ǫ) ≥ 1, and a1 , · · · , am ∈ Rd such that
|ak | = 1 (k = 1, · · · , m) and
d
{x ∈ R ; |x| = 1} ⊂
m
[
k=1
{x ∈ Rd ; ak · x > 1 − ǫ}
For any x ∈ Rd with x 6= 0, therefore,
x
ak ·
≥ 1−ǫ
1≤k≤m
|x|
max
This implies that
1
max (ak · x)
1 − ǫ 1≤k≤m
|x| ≤
In particular
x ∈ Rd
ak · B(t)
|B(t)|
1
lim sup max √
≤
1 − ǫ t→∞ 1≤k≤m 2t log log t
2t log log t
t→∞
1
ak · B(t)
1
=
max
lim sup √
a.s.
=
1 − ǫ 1≤k≤m
1−ǫ
2t log log t
t→∞
lim sup √
where the last step follows Theorem 5.1 and the fact that for each k, ak · B(t) is an
1-dimensional Brownian motion.
Finally, letting ǫ → 0+ on the right hand side leads to the requested upper bound.
5.11. Write
n
1
1X
1{Sk >0}
#{k ∈ {1, · · · , n}; Sk > 0} =
n
n
k=1
and
L{t ∈ [0, 1];
Sn∗ (t)
> 0} =
Z
1
1{Sn∗ (t)>0} dt =
0
n Z
X
k=1
k
n
k−1
n
1{S(k−1)+(nt−(k−1))Xk >0} dt
For each 1 ≤ k ≤ n, we consider two possibilities: First, Sk−1 Sk > 0. In this case
, nk ] and therefore
the continuous curve S(k −1)+(nt−(k −1))Xk has no zero point on [ k−1
n
1{Sk >0} and n
Z
k
n
k−1
n
1{S(k−1)+(nt−(k−1))Xk >0} dt
2
are either 0 at same time, or 1 at same time.
In the case Sk−1 Sk < 0, we use the bound
Z
1{S >0} − n
k
k
n
k−1
n
Summarizing our discuss,
1{S(k−1)+(nt−(k−1))Xk >0} dt ≤ 1
n
1X
1
1{Sk−1 Sk <0}
L{t ∈ [0, 1]; Sn∗ (t) > 0} − #{k ∈ {1, · · · , n}; Sk > 0} ≤
n
n
k=1
Thus, all we need is to show
n
1X
P{Sk−1 Sk < 0} = 0
n→∞ n
lim
k=1
Indeed, given M > 0, {Sk−1 Sk < 0} ⊂ {|Xk | ≥ M } ∪ {|Sk−1 | ≤ M }. Thus
n
n
k=1
k=1
1X
1X
P{Sk−1 Sk < 0} ≤ P{|X| ≥ M } +
P{|Sk−1 | ≤ M }
n
n
Given ǫ > 0,
√
lim sup P{|Sk−1 | ≤ M } ≤ lim P{|Sk−1 | ≤ ǫ n} = P{|U | ≤ ǫ}
k→∞
k→∞
where N ∼ N (0, σ 2 ), σ 2 is the variance of the i.i.d. sequence, and the last step follows
from the central limit theorem. Let ǫ → 0 on the right hand side:
lim P{|Sk−1 | ≤ M } = 0
k→∞
This leads to
n
1X
P{|Sk−1 | ≤ M } = 0
n→∞ n
lim
k=1
Thus,
n
1X
P{Sk−1 Sk < 0} ≤ P{|X| ≥ M }
lim sup
n→∞ n
k=1
Letting M → ∞ on the right hand side completes the proof.
5.15. (a) Write
Tn =
1
min{j; |S(j)| = n} and τ = {t ≥ 0; |B(t)| = 1}
n2
3
For any x > 0, write tn = [n2 x].
n
o
n
P{Tn ≤ x} = P min{j; |S(j)| = n} ≤ tn = P max |S(j)| ≥ n
j≤tn
o
By invariance principle, and a argument similar to Theorem 5.25,
1
d
√ max |S(S(j)| −→ max |B(t)|
t≤1
n j≤n
Notice that n ∼
√
x−1 tn (n → ∞). We have
n
o
n
o
n
o
lim P max |S(j)| ≥ n = P max |B(t)| ≥ x−1/2 = P max |B(t)| ≥ 1 = P{τ ≤ x}
n→∞
j≤tn
t≤1
t≤x
Summarizing our argument,
lim P{Tn ≤ x} = P{τ ≤ x}
n→∞
x>0
(b) We first claim that
∞
p
1 X
1{ρ̃i−1 ≤n} − 1{ρ̃i−1 ,ρ̃i ≤n} −→ 0
n2 i=1
(1)
Indeed,
1{ρ̃i−1 ≤n} − 1{ρ̃i−1 ,ρ̃i ≤n} = 1{ρ̃i−1 =n,
and
ρ̃i =n+1}
P{ρ̃i−1 = n, ρ̃i = n + 1} = P{ρ̃i−1 = n}P{ρ̃i = n + 1|ρ̃i−1 = n}
n+1
=
P{ρ̃i−1 = n}
2n
where the last step follows from (5.14).
In addition, for any i ≥ n + 1
P{ρ̃i−1 = n} =
X
(x1 ,···,xi−1 )
P{ρ̃1 = x1 , · · · , ρ̃i−1 = xi−1 }
where the summation is over all possible paths with x1 = 1 and xi−1 = n. By Lemma
5.31,
P{ρ̃1 = x1 , · · · , ρ̃i−1 = xi−1 }
= P{ρ̃0 = x1 }P{ρ̃1 = x2 , · · · , ρ̃i−1 = xi−2 |ρ̃0 = x1 } = n2−(i−2)
4
The number of the paths is at most
i−2
n−1
In summary,
∞
X
∞ X
i−2
2−(i−2)
P{ρ̃i−1 = n, ρ̃i = n + 1} ≤ n
n−1
i=n+1
i=n+1
∞ X
i−2
2−(i−n−1) = n2−(n−1) 2n = 2n
= n2−(n−1)
n−1
i=n+1
where the third step follows from Taylor expansion
∞ X
1
i−2
=
xi−n−1
n−1
(1 − x)n
i=n+1
|x| < 1
Hence,
∞
∞
1 X
1 X
2
≤
1
−
1
P{ρ̃i−1 = n, ρ̃i = n + 1} ≤ −→ 0
E
{ρ̃i−1 ≤n}
{ρ̃i−1 ,ρ̃i ≤n}
2
2
n
n
n
i=1
i=n+1
So we have (1).
We now claim that
∞
1 X
p
(τi − τi−1 − 1)1{ρ̃i−1 ≤n} −→ 0
2
n i=1
Write
Ln = max{j; ρ̃j = n}
We have that
∞
X
i=1
(τi − τi−1 − 1)1{ρ̃i−1 ≤n} =
Ln
X
i=1
(τi − τi−1 − 1)1{ρ̃i−1 ≤n}
Given ǫ > 0 and M > 0,
Ln
n X
o
2
P (τi − τi−1 − 1)1{ρ̃i−1 ≤n} ≥ n ǫ
i=1
2
≤ P{Ln ≥ n M } + P
n
k
X
o
(τi − τi−1 − 1)1{ρ̃i−1 ≤n} ≥ n2 ǫ
max 2 k≤M n
i=1
5
(2)
Let S(j) be a simple random walk and write γn = min{j; S(j) = n}. By Lemma
d
5.37, Ln = γn . Consequently,
n
o
2
2
P{Ln ≥ n M } = P{γn ≥ n M } = P max
S(j) ≤ n
2
j≤n M
n
1 o
1 o
√
√
√
−→ P max B(t) ≤
=P
max S(j) ≤
t≤1
M
M
n2 M j≤n2 M
1
n
as n → ∞, where the last step follow from invariance principle. Notice that the right hand
side can be made arbitrarily small by making M large. To establish (2), therefore, all we
need is to show
lim P
n→∞
n
k
X
o
(τi − τi−1 − 1)1{ρ̃i−1 ≤n} ≥ n2 ǫ = 0
max 2 k≤M n
(3)
i=1
Let n be fixed for a moment and write
Mk =
k
X
i=1
(τi − τi−1 − 1)1{ρ̃i−1 ≤n}
k = 1, 2, · · ·
We claim that Mk is a martingale with respect to the filtration {F + (τk )}. Indeed,
h
i
+
+
E Mk |F (τk−1 ) = Mk−1 + E (τk − τk−1 − 1)1{ρ̃k−1 ≤n} F (τk−1 )
and
h
i
i
h
+
+
E (τk − τk−1 − 1)1{ρ̃k−1 ≤n} F (τk−1 ) = 1{ρ̃k−1 ≤n} E (τk − τk−1 − 1)F (τk−1 ) = 0
where the last step follows from the proof of Lemma 5.32. By maximal inequality,
P
n
o
max 2 |Mk | ≥ ǫn2 ≤
k≤M n
1
ǫ2 n4
E max 2 |Mk |2 ≤
k≤M n
4
ǫ2 n4
2
2
E MM
n2
2
Mn
Mn
i2
4 X h
4 X
= 2 4
E (τi − τi−1 − 1)1{ρ̃i−1 ≤n} ≤ 2 4
E (τi − τi−1 − 1)2
ǫ n i=1
ǫ n i=1
≤
4
ǫ2 n4
M n2 C −→ 0
where the fifth step follows from Lemma 5.32. So (3) holds.
By (1) and (2), the problem is reduced to the proof of
Z ∞
∞
1 X
d
1{|W (t)|≤1} dt
(τi − τi−1 )1{ρ̃i−1 ≤n} −→
n2 i=1
0
6
(4)
By definition, ρ̃i−1 = ρ(τi−1 ) and |ρ(t) − ρ(τi−1 )| ≤ 1 for τi−1 ≤ t ≤ τi . Thus,
Z
τi
τi−1
1{ρ(t)≤n−1} dt ≤ (τi − τi−1 )1{ρ̃i−1 ≤n} ≤
Z
τi
1{ρ(t)≤n+1} dt
τi−1
i = 1, 2, · · ·
By the fact that ρ(t) = |W (t)|,
1
n2
Z
0
∞
Z ∞
∞
1
1 X
(τi − τi−1 )1{ρ̃i−1 ≤n} ≤ 2
1{|W (t)|≤n−1} dt ≤ 2
1{|W (t)|≤n+1} dt
n i=1
n 0
By Brownian scaling, for n > 1,
1
n2
Z
0
∞
(n ± 1)2
1{|W (t)|≤n±1} dt =
n2
d
This leads to (4).
7
Z
0
∞
1{|W (t)|≤1} dt